As first shown by Gorini, Kossakowski, Sudarshan and Lindblad given some linear map $\mathcal L:\mathbb C^{n\times n}\to\mathbb C^{n\times n}$, $e^{t\mathcal L}$ is a quantum channel for all $t\geq 0$ if and only if there exist $H$ Hermitian as well as $L_1,\ldots,L_m$, $m\in\mathbb N_0$ such that \begin{equation} \mathcal{L}(\rho) = -i[H,\rho]+\sum_{k=1}^m\Big( L_k \rho L_k^{\dagger} - \frac{1}{2} L_k^{\dagger}L_k \rho - \frac{1}{2}\rho L_k^{\dagger}L_k \Big)\tag1 \end{equation} for all $\rho$; see also this, this, or this post.
To pass to a more restricted version of this problem imagine we only knew that $e^{t\mathcal L}$ is a channel for certain, but not all times $t$. Note that this does not necessarily mean that $e^{t\mathcal L}$ fails to be a channel for some of the other times, just that we do not have any information about these times. In the limiting case where we're guaranteed to have CPTP for only one time $t_0$ (i.e. all we know is that $e^{t_0\mathcal L}$ is CPTP for some $t_0>0$) this does not automatically imply that $\mathcal L$ is of form (1). The reasoning is simple: take any bijective channel $\Phi$ (so one can define $\mathcal L:=\log\Phi$ using some branch of the complex logarithm) such that $\Phi$ has an odd number of negative eigenvalue—e.g., the qubit Holevo-Werner channel $\Phi:=\frac{{\rm tr}(\cdot){\bf1}+(\cdot)^T}3$ which has three positive and one negative eigenvalues. Then $e^{t\mathcal L}$ is a channel if and only if $t\in\mathbb N_0$, and for all other times the resulting map is not even Hermitian preserving: for all $t\in\mathbb R\setminus\mathbb N_0$ there is an odd number of complex eigenvalues which is at odds with the fact that complex eigenvalues of Hermitian-preserving maps always come in complex conjugate pairs. This begs the following question:
Given some $\mathcal L:\mathbb C^{n\times n}\to\mathbb C^{n\times n}$ linear and some subset $\mathcal T:=\{t_j\}_{j\in J}\subset(0,\infty)$ such that $e^{t_j\mathcal L}$ is completely positive and trace preserving under what assumptions on $\mathcal T$ can we conclude that $\mathcal L$ can be expressed as in Eq. (1)?
First observe the following sufficient condition: If $0\in\overline{\mathcal T}$, i.e. there exist times $t_j\in\mathcal T$ such that $t_j\to 0$ as $j\to\infty$, then $\mathcal L=\lim_{j\to\infty}\frac{e^{t_j\mathcal L}-{\rm id}}{t_j}$ is a limit of Lindblad generators${}^1$ $\frac1{t_j}(e^{t_j\mathcal L}-{\rm id})$, hence a Lindblad generator itself (as the latter, like every Lie wedge, form a closed set).
As seen previously it is not enough to know the channel property for just one time $t$. Another example—which will further motivate our question—is based on the example of a non-Hermitian matrix $A$ such that $e^{iA}$ is still unitary; there the corresponding generator $\mathcal L:=-i[A,\cdot]$ is not of Lindblad form (because not Hermitian-preserving) but $e^{\mathcal L}=e^{-iA}(\cdot)e^{iA}$ is a unitary channel. This example can be modified to the case where $\mathcal T$ is a finite, rationally dependent set. Interestingly, however, this example breaks down once we know that there exist $t_1,t_2\in\mathcal T$ rationally independent such that both $e^{iAt_1},e^{iAt_2}$ are unitary. In other words for the case of closed systems it is necessary and sufficient to have access to two (rationally independent) times.
Thus, in general, it is necessary that $\mathcal T$ contains two rationally independent elements if we want to conclude Lindblad form from the CPTP property; this is also in line with the initially discussed Holevo-Werner example. However, because the above example is tailored quite heavily to unitary matrices it is not clear to me whether this criterion is sufficient (beyond the special case of unitary channels); for example, already the implication that because $e^{iA}$ is unitary it is diagonalizable, $A$ must be diagonalizable does not have any counterpart for Lindbladians, cf. also the section "Quantum channels and dynamics" in the list of counterexamples in quantum information. In other words the following seems to be the simplest non-trivial—as well as a concise—version of my question:
Can it happen that, say, $e^{\mathcal L},e^{\sqrt2\mathcal L}$ are channels but the linear map $\mathcal L$ is still not of Lindblad form (1)?
1: Given any channel $\Phi$ it is known that $\Phi-{\rm id}$ (and any non-negative multiple thereof) is a Lindblad generator. This is also known as "Markovian approximation" (arXiv).
